= Solution
We take $\mathbb N=\{1,2,\ldots\}$ and require the <common difference> of an <arithmetic progression> to be positive. An <ultrafilter> $\mathcal U$ is a proper <filter on a set> deciding each <subset>: exactly one of $B$ and $\mathbb N\setminus B$ belongs to $\mathcal U$.
For the fixed length $m$, let $\mathcal B_m$ be the family of sets containing no m-term <arithmetic progression>. Any acceptable <ultrafilter> must contain every complement $\mathbb N\setminus B$ with $B\in\mathcal B_m$. These mandatory sets have the <finite intersection property>. Otherwise finitely many $B_1,\ldots,B_r\in\mathcal B_m$ would cover $\mathbb N$. Colour an <integer> by the first $i$ for which it belongs to $B_i$. The <Van der Waerden theorem> gives a <monochromatic> m-term <arithmetic progression>, contained in the corresponding $B_i$, a contradiction.
Generate a proper <filter> from those complements and extend it by the <ultrafilter lemma>. If some $B\in\mathcal U$ were in $\mathcal B_m$, its complement would also belong to $\mathcal U$, forcing the empty set into the <filter>. Thus
$$
\boxed{\exists\mathcal U_m\quad\forall B\in\mathcal U_m,\ B\text{ contains an }m\text{-term arithmetic progression}.}
$$
Notice that sets avoiding one fixed length need not be closed under finite unions; the argument uses the <finite intersection property>, not an unsupported ideal claim.
For the unbounded-length version, let $\mathcal B$ consist of sets whose <arithmetic progression> lengths are bounded. Each $B_i\in\mathcal B$ fails to contain a progression of some length $m_i$, and hence of every greater length. If finitely many $B_i$ covered $\mathbb N$, colour by the first containing index and apply the <Van der Waerden theorem> at length $M=\max_i m_i$. Again a contradiction results. The complements of all members of $\mathcal B$ therefore have the <finite intersection property>. Extending their generated <filter> gives an <ultrafilter with arithmetic-progression-rich members>:
$$
\boxed{\exists\mathcal U\quad\forall B\in\mathcal U\quad\forall m\ge1,\ B\text{ contains an }m\text{-term arithmetic progression}.}
$$
All <finite sets> lie in $\mathcal B$, so this <ultrafilter> contains every <cofinite set> and is a <nonprincipal ultrafilter>. The same is true of the fixed-length construction when $m\ge2$, because no singleton is allowed; for $m=1$ a <principal ultrafilter> also works. These constructions illustrate <ultrafilter selection from a partition-rich family>.
The infinite-length answer is \b[no]. Colour $n$ by $\lfloor\log_2n\rfloor\bmod2$, so successive intervals $[2^j,2^{j+1})$ alternate colours. An infinite <arithmetic progression> $a+td$, $t\ge0$, meets every sufficiently large such interval: its first term at least $2^j$ is less than $2^j+d$, which is at most $2^{j+1}$ once $2^j\ge d$. It therefore has terms of both colours. Neither colour class contains an infinite <arithmetic progression>, but every <ultrafilter> contains one of the two colour classes. This <dyadic block obstruction to infinite arithmetic progressions> rules out the proposed <ultrafilter>.
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